What is the extraneous solution to these equations? $\dfrac{x^2 - 33}{x + 7} = \dfrac{16x - 93}{x + 7}$
Multiply both sides by $x + 7$ $ \dfrac{x^2 - 33}{x + 7} (x + 7) = \dfrac{16x - 93}{x + 7} (x + 7)$ $ x^2 - 33 = 16x - 93$ Subtract $16x - 93$ from both sides: $ x^2 - 33 - (16x - 93) = 16x - 93 - (16x - 93)$ $ x^2 - 33 - 16x + 93 = 0$ $ x^2 + 60 - 16x = 0$ Factor the expression: $ (x - 10)(x - 6) = 0$ Therefore $x = 10$ or $x = 6$ The original expression is defined at $x = 10$ and $x = 6$, so there are no extraneous solutions.